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Question Number 81892 by Power last updated on 16/Feb/20

Answered by MJS last updated on 16/Feb/20

$$\mathrm{we}\mathrm{can}\mathrm{find}\mathrm{a}\mathrm{formula}\mathrm{for}\mathrm{any}\mathrm{value}\mathrm{of}\mathrm{the}$$$$\mathrm{next}\mathrm{number}\Rightarrow \mathrm{the}\mathrm{answer}\mathrm{is}\mathrm{not}\mathrm{unique}$$

Commented by mr W last updated on 16/Feb/20

$$nevertheless{he}^{\prime }llask‘‘solve{please}^{″}...$$

Answered by MJS last updated on 16/Feb/20

$$a_1=3,a_2=6$$$$\mathrm{\forall }n⩾3:a_n=a_{n-2}+a_{n-1}-n+2$$$$\Rightarrow n_5=17$$

Answered by MJS last updated on 16/Feb/20

$$a_n=\frac{5}{4}{n}^4-\frac{19}{12}{n}^3+\frac{91}{24}{n}^2-\frac{5}{12}n+1\Rightarrow a_5=26$$$$a_n=\frac{1}{4}{n}^4-2n^3+\frac{21}{4}{n}^2-\frac{5}{2}n+2\Rightarrow a_5=27$$$$a_n=\frac{7}{24}{n}^4-\frac{29}{12}{n}^3+\frac{161}{24}{n}^2-\frac{55}{12}n+3\Rightarrow a_5=28$$$$a_n=\frac{1}{3}{n}^4-\frac{17}{6}{n}^3+\frac{49}{6}{n}^2-\frac{20}{3}n+4\Rightarrow a_5=29$$

Answered by MJS last updated on 16/Feb/20

$$a_n=\frac{1}{2}{n}^3-\frac{7}{2}{n}^2+10n-4\Rightarrow a_5=21$$

Answered by MJS last updated on 16/Feb/20

$$a_n=\frac{16}{27}{\left(\frac{\sqrt[6]{108}}{2}\right)}^{n^3}{\left(\frac{16\sqrt{6}}{81}\right)}^{n^2}{\left(\frac{2187\sqrt[6]{2}}{256}\right)}^n\Rightarrow a_5=\frac{2187}{64}$$

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